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Let $R$ be a symmetric, transitive relation. If $(x, y) \in R$ then the symmetric property implies that $(y, x) \in R$. Using the the transitive property upon $(x, y)$ and $(y, x)$ we can conclude $(x, x) \in R$. Is this fair logic or is it flawed?

balddraz
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    Welcome to MSE. Please use descriptive titles. After reading the title of this question, users have no idea of which topic it belongs. – jjagmath Oct 14 '20 at 17:09
  • This is a common confusion. Showing $A \implies B$ does not mean you have shown $B$. You still need $A$ FIRST "to get at" $B$. In your case, you have IF $(x, y) \in R$, THEN $(x, x) \in R$. But you need $(x, y) \in R$ to begin with. – balddraz Oct 14 '20 at 18:05
  • This question is addressed in many posts already on the site: (https://math.stackexchange.com/questions/440/why-isnt-reflexivity-redundant-in-the-definition-of-equivalence-relation, https://math.stackexchange.com/questions/3802279/examples-and-counterexamples-of-relations-which-satisfy-certain-properties, https://math.stackexchange.com/questions/2106732/doesnt-symmetry-and-transitivity-imply-reflexitivity, etc.) Also, $\mathbb{R}$ is a standard symbol for the set of real numbers, so in general it would be better to use something else in this context. – halrankard2 Oct 14 '20 at 18:05

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You may conclude that for all $x$, if $(x, y)\in R$ for some $y$ then $(x,x)\in R$. That's not the same that proving that $(x,x) \in R$ for all $x$.

jjagmath
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